Find the multiplicity of eigenvalue $2a$

eigenvalues-eigenvectorskronecker productlinear algebramatricessymmetric matrices

Let $a,b$ be real numbers with $0<2a<b$. Let $K_n$ denote the commutation matrix of order $n$, and let $A$ be the $n^2\times n^2$ matrix defined by

$$A:=a[vec(I_n)vec(I_n)'+K_n-2diag(K_n)+I_{n^2}]+bdiag(K_n)$$

where $vec$ denotes the vectorization operator , and $diag(C)$ denotes the diagonal matrix having the same diagonal as $C$. Finally let $M$ be an $n\times n$ idempotent matrix with rank $r<n$, and let $B$ be the matrix defined by

$$B:=[M\otimes M]A[M \otimes M]$$

where $\otimes$ denotes the kronecker product. I am trying to find the multiplicity of eigenvalue $2a$ as a function of $M$. Also I would like to show that $2a$ is the smallest nonzero eigenvalue of $B$.

After some work I found that the eigenvalues of $A$ are as follows :

  • $b+na \, \text{ with multiplicity } 1 \, \text{and eigenvector} \sum_{i=1}^n (e_i\otimes e_i),$

  • $b \text{ with multiplicty } n-1 \, \text{and eigenvectors} \, \{(e_1\otimes e_1)-(e_i\otimes e_i), i=2,\dots, n\},$

  • $2a \, \text{ with multiplicty } n(n-1)/2 \, \text{and eigenvectors} \, \{(e_i\otimes e_j)+(e_j\otimes e_i), i\neq j\},$

  • $0 \, \text{ with multiplicty } n(n-1)/2 \, \text{and eigenvectors} \, \{(e_i\otimes e_j)-(e_j\otimes e_i), i \neq j\},$

where $e_i$ denotes the $n\times 1$ vector with $1$ in position $i$ and zeros elsewhere. I believe the multiplicity of $2a$ corresponds to the dimension of the $Col(M\otimes M) \cap Col(R)$, where $R$ denotes the $n^2\times n(n-1)/2$ matrix having columns $(e_i\otimes e_j)+(e_j\otimes e_i)$ for $i\neq j$, and $Col$ denotes the column space.

Any ideas how to proceed? Thanks a lot for your help.

Best Answer

By replacing $A$ by $\frac{1}{a}A$, we may assume that $a=1$ and $b>2$. With respect to the standard basis of $\mathbb R^{n\times n}$, the matrix $A$ represents the linear operator $f:\mathbb R^{n\times n}\to\mathbb R^{n\times n}$ defined by $$ f(X)=\operatorname{tr}(X)I_n+(X+X^T)+(b-2)\operatorname{diag}(X). $$ Hence $B$ represents the linear operator $g$ defined by \begin{align} g(X)&=Mf(MXM)M\\ &=M\left[\operatorname{tr}(MXM)I+MXM+(MXM)^T+(b-2)\operatorname{diag}(MXM)\right]M.\tag{0} \end{align} Observe that $g(X)=0$ when $X$ is skew-symmetric or when $MXM=0$. Hence $$ K=\{X\in\mathbb R^{n\times n}:X=-X^T\}+\{X\in\mathbb R^{n\times n}:MXM=0\}\subseteq\ker g. $$ As $g$ is self-adjoint (because $B$ is symmetric), in order to determine other eigenvectors or eigenvalues of $g$, it suffices to consider the restriction of $g$ on $$ K^\perp=\{X\in\mathbb R^{n\times n}:X=MYM \text{ for some symmetric matrix } Y\}. $$ Suppose $X\in K^\perp$. Then $MXM=M^2YM^2=MYM=X=X^T$. If this $X$ is an eigenvector $g$ corresponding to an eigenvalue $\lambda$, then $$ \lambda X=g(X) =\operatorname{tr}(X)M+2X+(b-2)M\operatorname{diag}(X)M.\tag{1} $$ It follows that \begin{align} \lambda\|X\|_F^2 &=\langle X,\lambda X\rangle\\ &=\langle X,\,\operatorname{tr}(X)M+2X+(b-2)M\operatorname{diag}(X)M\rangle\\ &=\operatorname{tr}(X)\langle X,M\rangle+2\|X\|_F^2+(b-2)\langle X,M\operatorname{diag}(X)M\rangle\\ &=\operatorname{tr}(X)\operatorname{tr}(XM)+2\|X\|_F^2+(b-2)\operatorname{tr}(XM\operatorname{diag}(X)M)\\ &=\operatorname{tr}(X)\operatorname{tr}(XM^2)+2\|X\|_F^2+(b-2)\operatorname{tr}(XM\operatorname{diag}(X)M)\\ &=\operatorname{tr}(X)\operatorname{tr}(MXM)+2\|X\|_F^2+(b-2)\operatorname{tr}(MXM\operatorname{diag}(X))\\ &=\operatorname{tr}(X)\operatorname{tr}(X)+2\|X\|_F^2+(b-2)\operatorname{tr}(X\operatorname{diag}(X))\\ &=\operatorname{tr}(X)^2+2\|X\|_F^2+(b-2)\sum_{i=1}^nx_{ii}^2\\ &\ge2\|X\|_F^2.\tag{2}\\ \end{align} Hence $g$ is positive definite on $K^\perp$, $\ker g=K$, every positive eigenvalue of $g$ is $\ge2$, and $$ \operatorname{rank}(B)=\operatorname{rank}(g)=\dim K^\perp=\frac{1}{2}m(m+1), $$ where $m$ denotes the rank of $M$. Moreover, $2$ is an eigenvalue of $g$ only if equality holds in $(2)$, i.e., only if $X$ is hollow ($\operatorname{diag}(X)=0$). This condition is also sufficient: if $$ X\in S=\left\{X\in K^\perp:\operatorname{diag}(X)=0\right\}\tag{3} $$ or equivalently, if $\operatorname{vec}(X)\in S_1=\operatorname{col}(M\otimes M)\cap\operatorname{col}(R)$ where $R$ is defined as in your question, then by $(0)$ we get $g(X)=2X$. Thus the multiplicity of the eigenvalue $2$ is equal to $\dim S\,(=\dim S_1)$.

This dimension can be zero. That is, $2$ is not necessarily an eigenvalue of $B$ or $g$. One obvious example is given by $M=0$, which makes $S=0$. For a non-trivial example, just try a numerical experiment that repeatedly samples some random orthogonal projectors $M\in\mathbb R^{3\times3}$ of rank $2$. Sometimes $2$ is an eigenvalue of $B$, but most of the time it isn't.

Since our problem formulation involves $\operatorname{diag}(X)$ or $\operatorname{diag}(MXM)$, it is not basis independent and we expect a nice answer when $M$ is, up to a permutation of rows and columns, in the form of $I_m\oplus0$ (where $m$ denotes the rank of $M$). Indeed, in this case one can easily see that the matrix subspace in $(3)$ consists of matrices of the form $F\oplus0$, where $F$ is any $m\times m$ hollow symmetric matrix. Therefore $2$ is an eigenvalue of multiplicity $\frac12m(m-1)$.

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